# A charge of -1 C is at the origin. How much energy would be applied to or released from a  -3 C charge if it is moved from  ( -1, 3 )  to (-9 ,7 ) ?

Feb 29, 2016

An energy of $6.16 \quad \text{GJ}$ should be added to the system to displace the $- 3 \text{C}$ charge from $\left(- 1 \text{m", +3"m}\right)$ to $\left(- 9 \text{m", +7"m}\right)$.

#### Explanation:

Electro-static Potential Energy: The electrostatic potential energy $U$ between two charges $Q$ and $q$ separated by a distance $r$ is given by the expression:

$U = - \frac{k Q q}{r}$,
where $k = 8.99 \times {10}^{9} \setminus \quad {\text{Nm"^2"C}}^{- 2}$ is Coulomb's constant.

Step 1: In this problem we have one charge (let us label it $Q$ ) fixed at the coordinate origin while the other one (labeled $q$) is displaced. The potential energy of the system changes when the distance between them change. To find the change in potential energy calculate the potential energies at the two points and subtract them.

If ${r}_{i}$ and ${r}_{f}$ are the distances of the initial and final points from the origin, then the change in potential energy due to the displacement is :

$\setminus \Delta {U}_{i \setminus \rightarrow f} = {U}_{f} - {U}_{i}$
$= \left(- \frac{k Q q}{{r}_{f}}\right) - \left(- \frac{k Q q}{{r}_{i}}\right) = k Q q \left(\frac{1}{r} _ i - \frac{1}{r} _ f\right)$

Step 2: Now evaluate $\setminus \Delta U$ numerically.

$Q = - 1 \text{C"; \qquad q = -3"C}$,

It is not mentioned what the units of the coordinates are. So assuming them to be in meters,

${r}_{i} = \setminus \sqrt{{\left(- 1\right)}^{2} + {3}^{3}} \setminus \quad \text{m"=\sqrt{10}\quad "m}$
${r}_{f} = \setminus \sqrt{{\left(- 9\right)}^{2} + {7}^{2}} \setminus \quad \text{m"=\sqrt{130}\quad "m}$

$\setminus \Delta U = k Q q \left(\frac{1}{d} _ i - \frac{1}{d} _ f\right)$
$= \left(8.99 \times {10}^{9} \text{Nm"^2"C"^{-2})(-1"C")(-3"C")(1/sqrt{10}-1/sqrt{130})1/{1"m}\right\}$
$= + 6.16 \setminus \times {10}^{9} \setminus \quad \text{J" = +6.16 \quad "GJ}$

So when the charge $q$ is moved from $\left(- 1 \text{m", 3"m}\right)$ to $\left(9 \text{m", 7"m}\right)$ the potential energy of the system increases by $6.16 \setminus \quad \text{GJ}$.

Step 3: Both charges are negative, so the interaction between them is repulsive. For repulsive interactions the potential energy increases if the charges are brought closer together and decreases if the charges are pushed farther apart. To determine if energy has been given to or released from the system, we must know the change in potential energy. If the potential energy decreases then the system releases energy. If the potential energy increases, energy must be given to the system.

Since we saw that the potential energy increased, energy must be added to the system to displace the charge.